Suppose the position of an object moving horizontally after t seconds is given by the following function \( \mathrm{s}=\mathrm{f}(\mathrm{t}) \), where s is measured in feet, with \( \mathrm{s}>0 \) corresponding to positions right of the origin. a. Graph the position function. b. Find and graph the velocity function. When is the object stationary, moving to the right, and moving to the left? c. Determine the velocity and acceleration of the object at \( \mathrm{t}=1 \). d . Determine the acceleration of the object when its velocity is zero. e. On what intervals is the speed increasing? \( \mathrm{f}(\mathrm{t})=\mathrm{t}^{2}-12 \mathrm{t} ; 0 \leq \mathrm{t} \leq 13 \) The acceleration of the object at \( \mathrm{t}=1 \) is \( 2 \mathrm{ft} / \mathrm{s}^{2} \). (Simplify your answer.) d . The acceleration of the object when its velocity is zero is \( 2 \mathrm{ft} / \mathrm{s}^{2} \). (Simplify your answer.) e. On what intervals is the speed increasing? The speed is increasing on the interval(s) (Simplify your answer. Type your answer in interval notation. Use a comma to separate answers as needed.)

**a. Graph of the Position Function** The position function is \[ f(t) = t^2 - 12t. \] This is a parabola opening upward. Its vertex is found by \[ t = -\frac{b}{2a} = -\frac{-12}{2(1)} = 6. \] At \( t=6 \), \[ f(6) = 6^2 - 12(6) = 36 - 72 = -36. \] The graph is a parabola that crosses the \( t \)-axis at the solutions of \[ t^2 - 12t = 0 \quad \Longrightarrow \quad t(t-12) = 0, \] so \( t=0 \) and \( t=12 \). (Note that the domain is \( 0 \le t \le 13 \).) --- **b. Velocity Function and Motion Direction** The velocity function is the derivative of the position function: \[ v(t) = f'(t) = \frac{d}{dt}(t^2 - 12t) = 2t - 12. \] - **Stationary:** The object is stationary when \( v(t)=0 \). Setting \[ 2t - 12 = 0 \quad \Longrightarrow \quad t = 6. \] - **Moving to the Right:** The object moves to the right when \( v(t) > 0 \). \[ 2t - 12 > 0 \quad \Longrightarrow \quad t > 6. \] - **Moving to the Left:** The object moves to the left when \( v(t) < 0 \). \[ 2t - 12 < 0 \quad \Longrightarrow \quad t < 6. \] Graphically, \( v(t)=2t-12 \) is a straight line crossing the \( t \)-axis at \( t=6 \). --- **c. Velocity and Acceleration at \( t = 1 \)** - The velocity at \( t=1 \) is \[ v(1) = 2(1) - 12 = -10 \text{ ft/s}. \] - The acceleration function is the derivative of the velocity: \[ a(t) = v'(t) = \frac{d}{dt}(2t-12) = 2 \text{ ft/s}^2. \] Therefore, at \( t=1 \), the acceleration is \[ a(1) = 2 \text{ ft/s}^2. \] --- **d. Acceleration When the Velocity is Zero** The velocity is zero at \( t=6 \). Since the acceleration is constant, we have \[ a(6)=2 \text{ ft/s}^2. \] --- **e. Intervals Where the Speed is Increasing** The speed is the absolute value of the velocity, \( |v(t)| \). Its rate of change depends on both \( v(t) \) and \( a(t) \). In one dimension, if \( v(t) \) and \( a(t) \) have the same sign, the speed is increasing. Since \[ v(t)=2t-12 \quad \text{and} \quad a(t)=2, \] we require \[ v(t) \cdot 2 > 0 \quad \Longrightarrow \quad v(t) > 0. \] Solve for \( t \): \[ 2t-12 > 0 \quad \Longrightarrow \quad t > 6. \] Thus, the speed is increasing on the interval \[ (6, 13]. \]